Lecture 11 : Implicit differentiation
For more on the graphs of functions vs. the graphs of general equations see Graphs of Functions
under Algebra/Precalculus Review on the class webpage. For more on graphing general equations,
see Coordinate Geometry.
The graph of an equation relating 2 variables x and y is just the set of all points in the plane which
satisfy the equation. We saw in Lecture 1 that in some equations relating x and y, we cannot solve for
y uniquely in terms of x. For example if we take the equation
x2 + y 2 = 25
and try to solve for y in terms of x, we get 2 new equations
√
√
y = 25 − x2
and
y = − 25 − x2 .
The graph of the equation x2 + y 2 = 25 is a circle centered at the origin (0, 0) with radius 5 and the
above two equations describe the upper and lower halves of the circle respectively.
6
x2 + y2 = 25
4
2
–5
5
–2
–4
6
4
4
2
f(x) = 25 – x2
2
–5
5
10
–5
–2
–4
5
10
–2
g(x) = – 25 – x2
–4
–6
–6
Now suppose we want to find the equation of the tangent to the circle at the point where x = 4 and
y = 3. One method of solving the problem would be to
√
√
1. Solve for y in terms of x, (getting 2 equations y = 25 − x2
and
y = − 25 − x2 .)
√
2. Decide which of these parts of the curve pass through the relevant point. (y = 25 − x2 )
1
3. Take the derivative of y with respect to x for the equation describing that part of the curve
−2x
(y 0 = 1/2 √25−x
2)
4. Calculate the value of y 0 when x = 4 giving us the slope of the tangent (y 0 = −4/3)
5. Find the equation of the line with that slope through the point (4, 3). (y − 3) = −4/3(x − 4) .
The above example was not difficult. However to apply the same method to find the tangent to the
curve:
2(x2 + y 2 )2 = 25(x2 − y 2 )
when x = 3 and y = 1 would be much more difficult. (See Notes at the end)
Implicit Differentiation
There is a much easier method (called implicit differentiation) for finding such tangents thanks to
the chain rule:
If y is defined implicitly as a function of x by an equation relating x and y, we treat y as a differentiable
function of x and proceed as follows:
Implicit Differentiation
1. Differentiate both sides of the equation with respect to x, treating y as a differentiable function
of x.
2. Collect the terms with y 0 (or
dy
)
dx
on one side of the equation and solve for y 0 .
Example Find the equation of a tangent to the circle x2 + y 2 = 25 when x = 4 and y = 3.
Note that we do not have to solve for y in terms of x and the calculations involved are much less
wearisome. In particular it applies to curves where solving for y in terms of x is very difficult. Both
approaches for the example given below are compared at the end of the notes:
Example
Find the equation of a tangent line to the curve described by the equation
2(x2 + y 2 )2 = 25(x2 − y 2 )
when x = 3 and y = 1.
2
When using implicit differentiation it is important to keep the following in mind:
dy
dy 2
dxy
= y0,
= 2yy 0 ,
= y + xy 0 ,
dx
dx
dx
where the middle identity follows from the chain rule and the one on the right follows from the product
rule.
dy
Example Find dx
= y 0 using implicit differentiation if y 5 + xy 4 = 2.
(Please attempt to solve this before looking at the solution on the next page)
√
√
d2 y
x + y = 1.
Example Find y 00 (or dx
2 ) using implicit differentiation if
(Please attempt to solve this before looking at the solution on the next page)
3
d √
d1
√
( x + y) =
dx
dx
1 −1/2 1 −1/2 dy
x
+ y
=0
2
2
dx
Solving for
dy
,
dx
we bring the terms with
dy
dx
to the left and all other terms to the right:
1 dy
1
=− √
√
2 y dx
2 x
√
multiplying both sides by 2 y, we get
√
√
2 y
y
dy
= − √ = −√ .
dx
2 x
x
To calculate y 00 =
d dy
( ),
dx dx
we have
" √ d(√y) √ d(√x) #
" √ 1 dy
#
√
√y x[ 2√y dx ] − y[ 2√1 x ]
−
x
y
d
dx
√
√ 2 dx
y 00 = −
=−
=−
dx
x
x
( x)
From above, we know that
dy
dx
√
y
= − √x . Substituting that into the expression for y 00 , we get
"√
y 00 = −
√
y
x[ 2√1 y [− √x ]] −
√
y[ 2√1 x ]
x
#
.
After cancellation and factoring −1/2 out of each term, we get
y 00 =
1
[1
2
−
x
√
y
√ ]
x
.
dy
Example Find dx
by implicit differentiation if y sin(x2 ) = x sin(y 2 ).
(Please attempt to solve this before looking at the solution on the next page)
4
Differentiating both sides with respect to x, we get
d
d
(y sin(x2 )) =
(x sin(y 2 ))
dx
dx
Using the product rule on both sides, we get
sin(x2 )
d(sin(x2 ))
dx
d(sin(y 2 ))
dy
+y
= sin(y 2 ) + x
dx
dx
dx
dx
Using the chain rule, we get
sin(x2 )
dy
dx2
dy 2
+ y cos(x2 )
= sin(y 2 ) + x cos(y 2 )
dx
dx
dx
Using the chain rule again, we get
sin(x2 )
To solve for
dy
,
dx
dy
dy
+ y cos(x2 )[2x] = sin(y 2 ) + x cos(y 2 )[2y] .
dx
dx
we bring all of the terms with
sin(x2 )
Factoring
dy
dx
dy
dx
to the left hand side to get
dy
dy
− 2xy cos(y 2 )
= sin(y 2 ) − 2xy cos(x2 )
dx
dx
out of every term on the left hand side, we get
dy
[sin(x2 ) − 2xy cos(y 2 )] = sin(y 2 ) − 2xy cos(x2 ).
dx
To solve for
dy
,
dx
we divide both sides by sin(x2 ) − 2xy cos(y 2 ), to get
dy
sin(y 2 ) − 2xy cos(x2 )
=
.
dx
sin(x2 ) − 2xy cos(y 2 )
5
Example The following curve is called the bouncing wagon. At what values of x does the graph have
horizontal tangents?
See full solution on next page. (Requires True Grit
6
).
We find y 0 and set it equal to 0.
2y 3 + y 2 − y 5 = x4 − 2x3 + x2
d
d 4
[2y 3 + y 2 − y 5 ] =
[x − 2x3 + x2 ]
dx
dx
6y 2 y 0 + 2yy 0 − 5y 4 y 0 = 4x3 − 6x2 + 2x
y 0 [6x2 + 2y − 5y 4 ] = 4x3 − 6x2 + 2x
y0 =
4x3 − 6x2 + 2x
.
6x2 + 2y − 5y 4
When y 0 = 0, we have the numerator must equal 0 and
4x3 − 6x2 + 2x = 0
or
x(4x2 − 6x + 2) = 0
so
p
√
36 − 4(4)2
6± 4
6±2
1
=
=
= 1 or .
x = 0 or x = ±
8
8
8
2
1
We have horizontal tangents at x = 0, x = 1 and x = 2 .
6±
7
Example (Not a good approach) Find the tangent to the curve:
2(x2 + y 2 )2 = 25(x2 − y 2 )
when x = 3 and y = 1. This curve describes a lemniscate, the graph is shown below.
To apply our method from page 1 of today’s notes (wrong way to go), we solve for y in terms of x. we
look at the equation as a quadratic equation in y 2 treating the x’s as constants,
2(y 4 ) + (4x2 + 25)y 2 + (2x4 − 25x2 ) = 0.
We get 4 equations for y:
s
y=±
−(4x2 + 25) ±
p
(4x2 + 25)2 − 8(2x4 − 25x2 )
.
4
Only two have graphs, since we cannot take the square root of negative numbers. The graphs are shown
below.
This is in fact an easy example, solving quartic polynomials (in y) is in general is difficult but there is
a formula. With powers of 5 and above, there is no general formula.
8
Example
Find the equation of a tangent line to the curve described by the equation
2(x2 + y 2 )2 = 25(x2 − y 2 )
when x = 3 and y = 1.
d
d
[2(x2 + y 2 )2 ] =
[25(x2 − y 2 )]
dx
dx
2
2
0
4(x + y )(2x + 2yy ) = 25(2x − 2yy 0 )
8(x2 + y 2 )x + 8(x2 + y 2 )yy 0 = 50x − 50yy 0
8(x2 + y 2 )x − 50x = −8(x2 + y 2 )yy 0 − 50yy 0
[8(x2 + y 2 ) − 50]x = −[8(x2 + y 2 ) + 50]yy 0
[8(x2 + y 2 ) − 50]x
= y0
−[8(x2 + y 2 ) + 50]y]
When x = 3 and y = 1, we get
[8(10) − 50]3
= y0
−[8(10) + 50]]
or
−9
90
=
= y0
−130
13
The equation of the tangent at this point is
(y − 1) =
−9
(x − 3).
13
9